Computing $\lim\limits_{n\to+\infty}\sum_{p=n+1}^{kn} \frac1 p$ for $k > 1$

Prove that

$\forall x>0,\quad \frac{1}{1+x}< \ln(1+x)-\ln(x)<\frac1 x.$

Deduce, for $k\in \mathbb{N}\setminus\{1\}$ ,

$\lim\limits_{n\to+\infty}\sum_{p=n+1}^{kn}\frac{1}{p}.$

We apply the mean value theorem to the function

t\mapsto \ln t

on the interval

[x,x+1]

. It exists

c\in (x,x+1)

such that

\ln(1+x)-\ln x= \frac1 c.

Since

x < c < x+1

, we have

\frac1{x+1}<\frac1 c<\frac1 x,

and we get the desired result. The bounds

\lim\limits_{n\to+\infty}\sum_{p=n+1}^{kn} \ln(p+1)-\ln p \le \lim\limits_{n\to+\infty}\sum_{p=n+1}^{kn}\frac1 p\le \lim\limits_{n\to+\infty}\sum_{p=n+1}^{kn} \ln p-\ln(p-1)

gives us by telescoping

\ln \frac{kn+1}{n+1}\le \lim\limits_{n\to+\infty}\sum_{p=n+1}^{kn}\frac1 p\le \ln k.

Finally, by the squeeze theorem

\lim\limits_{n\to+\infty}\sum_{p=n+1}^{kn} \frac1 p=\ln k.

P(X)\mapsto P(X+1)-P(X)